Problem: The lifespans of lions in a particular zoo are normally distributed. The average lion lives $13.2$ years; the standard deviation is $2.8$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lion living less than $18.8$ years.
Answer: $13.2$ $10.4$ $16$ $7.6$ $18.8$ $4.8$ $21.6$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $13.2$ years. We know the standard deviation is $2.8$ years, so one standard deviation below the mean is $10.4$ years and one standard deviation above the mean is $16$ years. Two standard deviations below the mean is $7.6$ years and two standard deviations above the mean is $18.8$ years. Three standard deviations below the mean is $4.8$ years and three standard deviations above the mean is $21.6$ years. We are interested in the probability of a lion living less than $18.8$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the lions will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the lions will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $7.6$ years and the other half $({2.5\%})$ will live longer than $18.8$ years. The probability of a particular lion living less than $18.8$ years is ${95\%} + {2.5\%}$, or $97.5\%$.